 Good, let's start then. Thank you for giving me half an hour extra. Sorry. No, I'll try to finish in the hour. So there was a very good question over the break, and this was something I was trying to swipe under the rug, and that's a bad idea. These correlation functions of what should I say? So in principle, the scaling dimensions of an operator in the car algebra, so this scaling dimension that you get after the twist, the way I gave it to you doesn't need to be an integer or a half integer. And that would mean that correlation if that's the case, that would mean that the correlation functions of the twisted shure operators, twisted translated shure operators, would have complicated branch cuts and not be single valued. This would be counterintuitive because these are just twisted correlation functions of four-dimensional theories where they are nice and single valued. So indeed, you can check explicitly by just going over the list of shure operators that such branch cuts don't happen. So the OPE in the car algebra, which is the twisted version of the four-dimensional OPE, is single valued. And so you don't get complicated branch cuts in the correlation functions of the car algebra operators. So now you have a structure that you can associate to every four-dimensional n equals 2 theories, namely that of a car algebra, a set of correlation functions that's completely miramorphic. And now you can basically ask two types of questions, the first one being, well, if you give me a theory, then what is the associated car algebra? What are the generators from which I can construct everything else? And what is their singular OPE? So this is one question. And you can answer this for a variety of questions. And we've tried to answer it at least for a variety of theories. We've tried to answer it in some sense. And so that's the first half of this lecture. That topic will be covered in the first half. And then you can ask a different question, which is sort of what is the space of all car algebra that come from a four-dimensional n equals 2 theories? And are there sort of lessons I can learn about four-dimensional n equals 2 theories that hold in full generality without referring to a specific theory that follow from this car algebra structure? So answering that question, or at least giving you an idea of what kind of lessons you can draw from just the existence of a car algebra will be the topic of the second half of the talk. So for the first half, car algebra, I already gave you one theory whose car algebra I wrote down. I wrote down its generators. And I wrote down the singular OPE. And that's it. And that theory was just a free hyper-multiple. The next thing to do is, of course, the theory of the car algebra of the free vector multiplet. And from that, we can build, hopefully, in some sense, the car algebra of gauge theories. So the first thing we should discuss is what are car algebra for gauge theories? And how does all of that work? So first of all, I need to give you the car algebra of the free vector multiplet. So the free vector multiplet sits in one of these representations that I discussed in some abstractly in a previous lecture. It sits in an epsilon 1, 0, 0, and an epsilon bar 1, 0, 0 complex conjugate. So that's the name of the free vector multiplet representation. I forgot what the modern way is referring to this representation. It's super conformal primary. It's phi, the same phi that Zohar wrote down in the previous lecture, which is just the scalar. If there's a non-privile non-abiding gauge algebra, it sits in the adjoint. Then, of course, I can act with a supercharge. I get a gegino. And here, I get the self-dual part or anti-self-dual part of the field strength. So this would be, I think, the epsilon multiplet. And the epsilon bar just points the other way and has the conjugate of everything. So which one is sure? There is a sure operator in this multiplet, which is good for us. This one is not because it is an asymmetry singlet. This one is also not because it's an asymmetry singlet. And in fact, this is the sure operator. So it's a gegino. It has j1 equals, in my case, j2 equals a half, j1 equals 0, r equals a half, and delta equals 3 halves. So I think that would give me a sure operator. And so this guy is sure with dimension h, which is delta plus j over 2. So that's one. So what kind of operator do we get? Well, if I define b of z to be the usual thing, which one do I take here? lambda tilde plus of zz bar in the q-columnology and the same z lambda i plus dot. Oh, here's a dot. Here's no dot, zz bar q-columnology. And this guy, I'm going to call not c, but del c of z. Then the OPE of these guys is 1 over z minus w, which just follows from the free theory. So computing the OPE is trivial. And the Kyra algebra is generated b del c with this OPE. So if I say del c, I can do this. But then I cannot really talk about c. And if you think about the multi-composition, like you do to get, for example, from the stress tensor self-OPE to the VeraZera algebra, what you see here is that the zero mode of c is not present in the Kyra algebra. So in fact, this space of states includes all the modes of b, all the modes of c, except the zero mode of c. So what I'm going to do is I'm going to say the Kyra algebra is generated by b and c, except the zero mode. And I'll add the little caveats to the equations that are to follow. But unless you force me to, I was not planning on explaining this any further. The details are in our paper. It's a small subtlety. So it's kind of a technicality. And for that reason, I don't think it's particularly conceptually very important for what I'm about to discuss. So this is a bit of a silly choice. I should have just called this something like d, right? And then we would have a bd. And then it would all be fine. And then I wouldn't have to talk about this zero mode. But this silly choice will be useful for what follows. And that's why I'm doing this. So now let's talk about non-trivial gauge theories. We're going to be nice and abstract. So let's not commit to any specific theory. So we're going to define a general gauge theory, t gauge as three vectors, well, as vectors, vector multiplets in the adjoint of some gauge algebra. Don't need to write that down. G coupled to, well, in principle, if you have a Lagrangian theory, this would be hypermultiplets. But of course, I can take some crazy non-Lagrangian theories. And all I require here is it for this symmetry to be gaugeable at this level is that it has a global symmetry g that I can gauge. So I'll take a general t with global symmetry g. For example, a bunch of free hypers would do the job often. But I could take another theory. And then I couple them in the usual way. I gauge the symmetry. And then, of course, I compute the one loop beta function. And if I want this to be a super conformal field theory, then I require that the beta function vanishes at one loop, and by n equals 2 super symmetry, that's sufficient for complete conformality. And the one loop beta function vanishing is the same as saying that the level of the flavor central charge of this theory, k4d, is equal to 4 times the dual coccitor number. So fairly abstract, but I hope it's clear what we're doing here, right? So a question, natural question to ask now. If you give me not this theory t, but just its chiral algebra, and you give me a bunch of vectors in the adjoint, well, instead of the full vector multiplet, if you give me the algebra associated, the chiral algebra associated to these vector multiplets, can I do the gauging purely at the level of the algebra? Is there some operation that I can do if you just give me these two chiral algebras that gives me the chiral algebra of the gauged theory? So for example, the chiral algebra here has gauged non-singlets, right? In principle, I don't have to. So in principle here, there are operators charged under g, and same here. And of course, in the end, I want to only have gauged shinglets in the chiral algebra. So what's the operation? Is there an operation that I can do that sort of combines these two chiral algebras in some non-trivial way and gives me the chiral algebra of the gauged theory? And the answer to this question is yes. And it's a very nice prescription. So let me use this notation to say the chiral algebra of TG from the chiral algebra of T, which in particular contains a g-affin-cos-moody current at level k2d equals minus 2 times to do a coaxial number. So in particular, so what I'm saying here is just that this chiral algebra of this original theory contains some current, ga of z, where is an adjoint index of the free vector multiplets. Whoops, the kind of algebra of the free vector multiplets. We know it's generated by BACA, modulo of the zero amount. And the way to do this, so the prescription is as follows, I'll define it's a BRST called homology prescription. So you define the BRST charge as the integral dz over 2 pi i of the BRST current, JBRST of z. So this counter integral, if you act with the charge on a given operator, you take this counter integral around the operator. And all it does, of course, is just pick out the simple pole in the OPE of this BRST current with the operator. So alternatively, you can just look at the OPE, take out the singular, the simple pole in the OPE, and that's what the action of the BRST operator is. Where JBRST is the usual one. So we see. Well, here it's fine. And the claim is now that the carrier algebra of the gauge theory is the BRST called homology. So it's, again, a homological problem. We're already in the homology of funny cube, but now we do a BRST homology in the carrier algebra. So it's a BRST homology over the space of states that are either in chi t or the carrier algebra of the three vectors. And this is true. Margello the zero mode issue. Oh, sorry. The Margello has to go outside. So this is important. So I say BRST homology, but everyone to take the homology of this guy, I better check that it's no potent. So I have to do something like this. QBRST on anything is 0. And indeed, this is true. So note that this is true only if. So there's a current here which has a non-trivial OPE, but itself OPE is determined, the singular term is determined in terms of this level. And only if k4d is equal to 4 times the dual cox to the number of g. So if the Monlou beta function vanishes, then I can do this, play this game. And if not, I cannot. So this is the claim. I'm not going to do this in very much detail, but as a first. So a zero order consistency check that you can do is check that if an operator is BRST closed, then it's automatically gage invariant. This is a true fact of life that I wasn't going to explain. But at the very least, restricting yourself to the QBIRST closed guys indeed kills all the non-gage invariant states. In fact, that one property may be something I can actually show. So how am I doing? Yeah, I think I have time to do that. So let me do the example of a charged operator, or i of z, in the Chiara algebra. So we have this Chiara algebra, as I said, contains an affrancatsmoody algebra. It may be much bigger. It may contain all kinds of charged operators. Let me just take one of these operators and check that it's not BRST closed. The OPE of a charged operator, basically, by word identities, has to take this form. So it can have more singular terms. It can have more regular terms, but this is the OPE of the current with any charged operator, where this is, of course, the associated representation matrices. So now it's easy, because the action of the BRST operator this guy, well, let me insert it at the origin for simplicity, is just a simple poll on this OPE. So there is some CA. But CA comes from another sector. It comes from the free vector multiplet. So the OPE of c with this guy and b is automatically regular, because these sectors don't talk to each other. So I have CA, A, G, A minus 1 half, and then FABC, CA, where did I put the C, CB, BC. And the only thing that's going to give me a simple poll is precisely this OPE. So this is CA of 0 TA. And that's it. So here, you see some non-trivial composite operator. It's the normal-ordered product of these two guys, but it's certainly not 0. So this is not BRST closed. Lambda becomes B or C. And so once you can do it as an exercise, try to see that your B and C operators are also not too close. So here, it was important that I took a charged operator that was in the kera-algebra of T and not in the other kera-algebra, the kera-algebra of the free vector, but you can easily show that the free, the keginos also disappear. And so in the paper, we have a bit more explanation than I have time to give. In particular, we can think of this BRST operator as being something like a one-loop correction, morally speaking, at least, to our funny supercharge. So we're in the homology of funny queue. And now we're going, in that homology, restrict to a further homology. And in some sense, this is just a one-loop correction to the supercharge that you would get if you would properly define it in the interacting theory. We also have a claim that this is enough, that this one-loop correction, just like the beta function, if you check it at one loop that it vanishes, it vanishes completely. Our claim is, we have a similar claim that this one-loop correction, this BRST game, is indeed enough to get the kera-algebra of the interacting theory. Correction in the sense of bijection? If you want, yes. It's the one-loop correction. Yeah, I don't have an explicit check of that, but it's really the one-loop correction that you would get if you would write down the interacting Lagrangian and properly determine the associated Q and S out of which our funny queue is made. So this is a nice trick that allows you to, in some sense, quickly determine the kera-algebra of gauge theories. Maybe I should jump ahead a little bit to change my order a little bit from what I wrote down. So I kept one thing of Zohar's notes, of Zohar's lectures. Let's look at this theory, SUN with two n-hypers in the fundamental. And let me take n greater than 3, because for n equals 2, the flavor symmetry gets enhanced. n equals 2, the flavor symmetry gets enhanced to SOH, and that enhancement gives you a more fun kera-algebra, but not something I want to talk about in detail now. So what are your, what operators survive this BRST game and are part of the kera-algebra? Well, they have to be gauge invariants. So first of all, let's write down what we have in the un-gauge theory. So in the un-gauge theory, or the free theory, we would have free-hypers. We have q and q tilde. And these, of course, have our symmetry indices as usual. They have flavor indices, A, A, and they have gauge indices, I, I. And actually, I'll do it the other way around. So this will be flavor, and this will be gauge. So this gives rise to a pair of simplectics bosons we've seen before for each A and I, q tilde, A, I. And then there is the free vector. So we have to get genus, and those give rise to your BC system. And out of this, we have to build gauge invariant combinations. And not only gauge invariant combinations, we need to build combinations such that they are BRST closed. And in fact, there's a very nice, there's a very natural thing you can do, which is just this. So you take q, where did I put the indices, q, A, I. So this is no tilde, q, A. So just take the two hypers and you contract their gauge indices. And if you symmetrize this, so you get an operator of dimension two, which is a triplet of the r symmetry. So this is the super conformal primary of a B hat one multiplet that I discussed before. And in the Kyra algebra, so this gives rise. So this is in the B hat one multiplet. And in the Kyra algebra, this is of course the same thing as q, A, J, q, A, I, which I define to be some dimension one operator. So remember that these were dimension one half, q, q tilde were dimension one half. And so this is a dimension one operator in the Kyra algebra. And this is precisely my flavor symmetry current. So now you can see this sort of in two ways. You can view this flavor symmetry current as coming from a B hat one multiplet, which you had in that theory, or you can see it as just normal ordering the elementary building blocks in the theory. And in fact, you can split it into a trace. So the trace gives you the U one current and the rest gives you the SU and F current. You can check that this is an U one. I think I wrote down the level somewhere. Yeah, so this one doesn't really have a level because the normalization is ambiguous because there are no structure constant to normalize the thing by. But this one has a level, but you can easily determine because you know the OPE of the elementary building blocks. So you can just take the OPE of this current with itself and you find that it has minus Nc. So here, sorry, I work in conventions where this is Nc and two Nc. This is equal to Nf. So this is how you can sort of start building your Kyra algebra. You have the engaged non-invariant building blocks. You combine them in a gauge invariant way. And then of course you check that it's BRST closed and not BRST exact. And then you get operators in the gauge theory. So these are simple ones. There's of course also the C hat 0, 0, 0, the stress tensor multiplet, there's the SU2R symmetry current that we already saw before, Ij. So in the 4D theory upstairs, it's some complicated combination. In the Kyra algebra, it looks a bit simpler. It looks like this in terms of the elementary building blocks. Q, let me not contract all the indices. They're contracted in the obvious way, minus lambda, minus b dash c. So this, you can check that this particular combination is actually in the columnology of Q, BRST. And it's sort of precisely what you would get if you would take the four-dimensional expression for the SU2R symmetry current and do the twist. Because if you take the SU2R symmetry current, you basically see all the fields that are charged under SU2R and in the free hyper, it's precisely this and in the free vector, it's precisely this. And you don't see the gauge field, you don't see the scalar dimension one scalar in the vector multiplet that you don't see the fermions in the vector multiplet because they're not charged under the SU2R symmetry. So there's a funny thing happening here which I'd like to point out, which is that if you have a flavor symmetry current, I can always define the so-called Sugavara stress tensor, T-Sugavara, which acts, this guy acts as a stress tensor on the currents, but it may not act as a stress tensor in everything else. So the Sugavara stress tensor is just some dimension two operator in the Kyberaw algebra, which I can build from if I have currents. And this is a proper thing I can always do. As it happens, in this case, T of Z minus the Sugavara stress tensor of Z. So in the original Kyberaw algebra, there are certainly two different operators. You can just look at how they're constructed. This T of Z is built out of Q's, Q tilde's, B's, and C's, whereas this J is built out of two Q's, so this guy is built out of four Q's. So these are different-looking operators. And in principle, they're different in the Kyberaw algebra. However, if you take their difference, then you find that this is Brst exact. And so in the gauge Kyberaw algebra, these two operators become the same. So well, now we have to make some conjecture as to what the full Kyberaw algebra of this theory is. So we have some generators that we knew about. We know that, well, there's these currents. Then we thought maybe the stress tensor is a new generator, because I cannot build it as a normal-ordered product of currents. But that's false, because the stress tensor is actually the same in the gauge theory as a normal-ordered product of currents. So we don't have that as an extra generator. And so maybe we're done. Maybe this is just the full Kyberaw algebra, right? So I'm not going to write the current self OP again. We've seen it before. So I can just claim, well, maybe these currents generate the Kyberaw algebra, and their self OP is the canonical one for an FI cut smoothie current. Is this true? This is a tough question at this level. Are there natural other gauge invariant operators that I can build out of, let's say, the hyper multiplets? Sorry? Variants. Thank you. And it turns out that these are new generators. So we have new generators. These turn out to be generators, which are q, q, and then some epsilon to contract all the gauge indices. So where did I put my gauge indices? A1, An, A1, An, and then I1, In. So they're in the enfold anti-symmetric representation of the flavor index. And in the Kyberaw algebra, I just make the big q's, little q's. Sorry, I should have done that right away. And this defines some baryon operator. So in the ungauge theory, this was not a generator. Because in the ungauge theory, I had the q's to play with. And if the q's are generators, then I can just build this guy out of q's. But in the gauge theory, the q's have disappeared. They're no longer in the columnology. And so I need to add these guys to the Kyberaw algebra. There's new generators. And so in this way, the game of determining what your generators are becomes very non-trivial. And of course, there's also b tilde. From gauge theory, I'm going to switch to q. From a 2D point of view, you could actually not have guessed that. Because there is no need. This algebra, which was just the u1 plus SUNF afrenkatz moody symmetry, closes on itself. So it's a consistent subsector, as it happens, of the full Kyberaw algebra of this theory. And so there was no way, if someone would give you this, there was no way to check that it was inconsistent as a Kyberaw algebra. But that's not what we're asking. What we're asking is, what's the full Kyberaw algebra of this theory? And then we have to deal with these kind of issues. So our conjecture is that this is it. So we have the Kyberaw algebra generated by this guy and the Kyberaw algebra. And then the two baryons, it's baryonic type operators as additional generators. And our conjecture is that's it. So this is beautifully stated in the grandian description, but you want to go beyond, right? I want to go beyond. Yes. Are you saying I don't have time to go beyond? No, no. I am asking how you know if there are baryons or more crazy things beyond? Oh, if you, OK, let's ask about beyond Lagrangian a little bit later. So for Lagrangian theories, now this conjecture stands and it's not proven. In fact, even for Lagrangian theories, as soon as they're noninteracting, this BRST problem becomes non-trivial. And we have no proof for the full Kyberaw algebra of any Lagrangian theory. Even here, the simplest Lagrangian theory you can write down, basically, we don't have a proof that this is the full set of operators in the BRST homology of the Kyberaw algebra that started with these gauge non-invariant fields. So this is purely a Kyberaw algebra problem at this level. It's just a BRST homology problem. And we were not able to solve it. No, the bootstrap analysis, you can do small things. Like you can try to take a self-OP of this guy with itself. And maybe you see some crazy object that you could not generate in any other way. Then that would mean, well, you add this object to it. But at some point, we often find this, we check that this doesn't happen. And that sort of the Jacobi's are satisfied. So this is, again, a nice close subalgebra. But are there other generators? I don't know. In fact, we don't even know if, for any theory, we don't know for sure if the Kyberaw algebra is finitely generated. Could have infinitely many generators. I don't believe that to be true. I strongly believe that they are finitely generated. But they don't have a proof for it. And you see how complicated the question of generators becomes in this BRST homology problem. You throw away some stuff. And so suddenly further down, there could be things that were not generators before that suddenly become generators. Sorry? What kind of Kyberaw algebra does the value of the generator? I don't think the Kyberaw algebras, the question is what Kyberaw algebra do the variants generate. And they cannot be viewed independently from the currents as it happens. Because in the self-OPE of the variant, anti-variant, you will find the currents back. So this does not close on itself as a Kyberaw algebra. So it's just some extra sector that needs this Kyberaw algebra. I lost my board. Gravity seems to work in mysterious ways here. So what should I say? I should say a few things very quickly in words. One thing I want to say is if you want to check these kind of conjectures, you can sort of go over the list of all the operators that you can build at low orders. You just build all the operators that you can from these building blocks. And you'd solve your BRST problem order by order in this dimension. And then for the first few levels, you find no new generators. And you count yourself lucky you think, OK, maybe my conjecture is true. Then in principle, there's another thing that you can do, which is compute the super conformal index. There's a particular limit of the super conformal index, the so-called SURE limit. So this is essentially the S3 cross S1 partition function, supersymmetric partition function, which in a particular limit just counts to SURE operators. This is a happy coincidence. It counts them with signs. So it becomes almost a character of the Kyberaw algebra, except there's a minus 1 to the f. So it's a graded character of the Kyberaw algebra that you get from the SURE limit of the super conformal index. And the Kyberaw algebra itself has no supersymmetry. So I can't say that this is the index of a Kyberaw algebra. That's a meaningless term at this point. So it's just a graded character of the Kyberaw algebra. And so you can, for example, if you have a conjecture for a Kyberaw algebra, you can compute its graded character and see that it matches, whether it matches or not, the super conformal index of the four-dimensional theory. This gives you some evidence, but not full evidence, because there could be cancellations. You don't know the exact structure of nulls, et cetera. But it would help you a lot. Unfortunately, as it happens, computing the super conformal index for these theories is relatively easy. Computing just for this Kyberaw algebra, computing for this guy with these generators, computing the graded character, or the character of this thing, is very difficult. That's because these are affine currents at negative level, for which the usual cuts formalized for the character don't work. And so basically, we can do this only level by level. And so we can go down to level six or seven, and then our mathematical code gives up. And so that's the order to which we check these conjectures. So there's an indexology section that I will completely skip. I can say much more about the connections to the index, the various limits, et cetera, et cetera. But I want to get to other stuff. So let me focus on that instead. So before I end this section on Lagrangian theories or specific theories, I want to add some conjectures for Kyberaw algebra of specific theories. All of these are conjectures. So nf equals 2nc, super conformal QCD. That's the one we just discussed. So it's SU nf plus U1 at level minus n affine cuts moody symmetry. So that's the current part plus variance with OPEs that you can find in our paper, or at least some case. For similarly, there's a nice so-called Minahan-Nemeshansky theory. So let's look at the E6 theory. This theory is a bit obscure. It's not Lagrangian. So there's no gauge theory description like this. But the index is known. And we know that it has E6 flavor symmetry. And so at the very least, it has E6 affine cuts moody symmetry. And as it happens, the level turns out to be minus 3. So we also know that about the theory. Our conjecture is that this is all there is. There is no other generator in the Kavara algebra. This is not trivial. We checked it again against the index. But otherwise, we can check it, because it's just some abstract theory that we don't know very little about. So let me also say that if your theory has extra supersymmetry, for example, n equals 4 super young mills. What do you mean by n equals 4 super young mills? Those I don't know. I'm not so sure about that. For sure, contain this. But I'm not sure if that's the entire Kavara algebra. n equals 4 super young mills. So the Kavara algebra of n equals 4 super young mills, let me just say this, has 0,4 Susie. Well, it's Kavara, so I can just say it as n equals 4 super super young mills. And I should add the qualifier if you want to be precise that this is small n equals 4. So this just happens because you have extra supercharges floating around in the four-dimensional theory that to revive the twist, and they become super, or super currents, they become super currents in the Kavara algebra also. And the Kavara algebra of n equals 3 theories has n equals 2 supersymmetry. And I should say that these are super VRZera algebra. So just like the flavor symmetry gets enhanced to this infinite-dimensional Katsumudi algebra, these supersymmetries get enhanced to infinite-dimensional super VRZera algebras. So we have some conjectures for these Kavara algebras. But in the interest of time, I was not planning to discuss them. And so you can, of course, continue this list forever. And for all your favorite theories, you can try to determine what the Kavara algebra is. Try to solve the cosmology problem, or try to find it in some different ways. I think that would be interesting. What would be extremely interesting would be to try and see if you can, at least in one or two cases, solve this cosmological problem and have a conclusive proof for the Kavara algebra of the corresponding theory. OK. Any questions about Kavara algebras of specific theories or other questions you've had? So what's TN? TN is very nice. It has, of course, the three flavor symmetries. And those sit at, actually, the critical level. Because remember to gauge a theory, I need k2d equals minus 2h. But I gauge two TNs together, or two punctures together. And so for TN, it's minus the dual-coxater number, which is a very nice level to be at for a flavor symmetry. In particular, the Sugovara stress tensor, if you know what that is, it becomes a null state. Oh, I just introduced it, so you know what it is. It becomes null in itself. So in these three Afan-Kazmoudi algebras, there is no Sugovara stress tensor. But of course, the full TN theory has a stress tensor still. And then we believe there are some other generators floating around. This is not the full thing. But we don't have a good. We have some ideas about it. But we don't have, like for the others, we don't have a proof. Yeah, and then, of course, you can close punctures. And that's another beautiful procedure. Gaging was beautiful in the sense that you can do it purely at the level of the Kera algebra. If you know about the N theories, you also know that you can close punctures. This corresponds, you can also do this purely at the level of the Kera algebra. It's a procedure called Drinfeld-Sokolov reduction. And it's also be a risky problem. But it has to kill some of the flavors in it. For general class S theory, that's a huge class to us. I mean, we're just like simple gauge theories. We don't know. So for general class S, there's, of course, now a picture, which I can sketch, which is just you gauge and you close punctures and you do Drinfeld-Sokolov reduction. But for general class S theory, it's, yeah. It depends on the puncture, depends on the puncture. It depends on? It depends on the puncture composition. Yes. Oh, that's nice. So you want to ask if this is S2 or manifestly. That's correct. Gaging depends on the puncture composition. And it's non-trivial that if you take two different puncture compositions, you get the same Kera algebra. So this, in some sense, you could hope, would be some kind of bootstrap for Kera algebras, right? I take the Kera algebra of Tn and I glue it together in two different ways. And then I should get the same Kera algebra. So is there some constraints on the original Kera algebra that they can find in this way? I don't know the answer to that question, but I think it's fascinating. It would be very nice if you can actually get somewhere there. I should also say that basically the reason, when I claim that this PRSK problem sort of solves the whole thing, it also means there's no further coupling dependence. So you can actually show that all the three-point functions in a gauge theory are independent of the coupling. So all of this is sort of fixed by free theory OPs. There's no further coupling dependence. OK, I'm running out of time, but I have 10 minutes for two pages, which is going to be great. Let's do non Lagrangian theories just so that I can show you that there are certain bonds you can get, certain uniterity bonds that follow from this Kera algebra story. So let me take a theory T and let me suppose that it has flavor symmetry, flavor symmetry algebra G. Then if T is a good theory, there is an associated current. And in the Kera algebra, we have delta AB plus I of ABC, FC of JC of 0 divided by Z. Of course, if it's a good theory, then it also has a stress tensor. And if the theory has a stress tensor, then let me remind you it has an SE2R symmetry current in the same supermic conformal multiple. And that guy gives me a stress tensor in the Kera algebra, so with the usual OPE. And finally, if this guy is a good stress tensor, which we'll assume, then the stress tensor OPE with this current is, looks like this. Oh, this is it. So this just, you can either compute it upstairs or just follow us from the fact that this is a stress tensor and this guy is a free or zero primary. The fact that it's a free or zero primary follows because there can be nothing of even lower dimension. So this is a consistent subalgebra. The OPE closes here. So in principle, this is all perfectly good. And it, of course, determines all the JT correlators. So this determines the TJ correlators. And also the correlation functions of their normal order products. So for example, you don't need to copy this, but I am going to need it a bit later. So this is in our paper also. Let me take a four point function. And let me move the four points to the canonical positions, point one at the origin, point two at some coordinate z, point three at one, and point four at infinity. And then this has to, as z goes to zero, it has prescribed singularities. As z goes to one, it has prescribed singularities. And as z goes to infinity, it also has prescribed singularities. So these singularities come from this. And so you can just solve this. It's easier to do in Mathematica, but what you get is something that looks like this. z squared over 1 minus z squared delta AD delta BC. So these sort of take into account the three strongest singularities as you go to 0, 1, and infinity. And then there are some subleading singularities, k2d z minus 1. So this four point function is just fixed. It follows directly from this simple equation. So now let me look at a particular operator. Let me call it j squared of z. And I define it to be jA, the normal ordered product of jA of z, jA of 0. So in other words, it's just the limit as w goes to z of jA w jA of z minus the singular terms, k2d over z squared times delta AA, which is just a dimension of g. And that's it. The subleading term here, this is also, in principle, a singular term. But of course, by the anti-symmetry of the structure constants, it vanishes if I contract the indices in this particular way. So this is how I define the operator. And this should be w minus it. So what is this operator? It's some operator in the Kaira algebra. Where does it come from? It's a dimension two singlet. So it's basically our Sugavara guy. But where does it come from in four dimension? It could be the stress tensor. It could be the same operator as the stress tensor here. And then it would come from a C hat multiplet, a C hat 0, 0 multiplet. There's another option. If you just go over the list of possible multiplets and you have to, and you see what this j squared of z could be, given also the fact that it appears in the OPE of j with itself, it can be C hat 0, 0 or a B hat 2 multiplet. And that's about it. I think it can also be a higher spin current, but let's assume our theory doesn't have higher spin symmetry. And then it can be either this or that or a combination of these two, so linear combination. What I'm interested in is the B hat 2 part of this linear combination. In general, this guy will be a linear combination of the two. And I'm interested in the B hat 2 part of the guy. So how do I extract it? Well, a B hat 2 will have 0 two point function with the C hat 0, because these are different types of operators. And the C hat 0 I know, because the C hat 0 is just a stress tensor. So I'm going to make sure I'm going to add to this j squared a multiple of the stress tensor so that it's orthogonal to the stress tensor itself. Are we good? So let's write all the norms, or in this case, two point functions for all the operators we're considering. So these are the dimension two operators. t z t 0 is this. t z, sorry, t z j squared of 0. Well, that's just the limit as that goes to w, or as that w goes to 0, in this case, of t of z, j a of w, j a of 0. And this three point function I can compute from the OP of t with itself, with j. So it's a limit as w goes to 0 of, well, let me do this schematically. So j a of w times j a of 0 over z squared plus del j a over z plus dot, dot, dot. And then plus the other one where you swap w with 0. And then you can use the fact that by cluster decomposition, this guy should go to 0 as z goes to infinity. So this vanished by, in the total sum, these will vanish by cluster decomposition, because this is just a dimension two operator. So I only need to look at the singular terms. The singular terms are a little bit non-trivial to work out. So you get some funny cancellations. But in the end, you find that this is something which I have, I do have it. It's k 2d times the dimension of g. That comes from the two-point function of jj, contracted with itself over z to the fourth also. And then, of course, there's j squared of 0 with itself. But that is very simple if I realize that that's just a limit of that four-point function. It's the z to 0 limit with all the singular terms subtracted. So this is immediately computable. I don't have time to do it exactly. But I think it's 2 times k 2d times the dimension of g again. Did I do this correctly? Yeah. Dimension of g times k 2d plus the dual-coxeter number. Where the dual-coxeter number is something like f abc, f abc, up to factors that I forgot. But it's something like that. So you see, my two-point function j squared is not orthogonal to the stress tensor. My operator j squared is not orthogonal to the stress tensor because this two-point function is 0. So we're going to define the new guy, j hat of z. And I want it to be orthogonal to the stress tensor. So to do that, I'll take it to be j squared of z minus whatever this thing is. So it's 2k 2d dimension over c times t. So this, I claim, has to be a b hat 2 because t is by construction the c hat 0 is 0. And you also have to check that it is not a descendant. But in this case, you can show that it's a primary. And so it has to come from a b hat 2. So j hat of z is a twisted b hat 2 operator. B hat 2, a twisted b hat 2, well, so more precise. I mean, let me write operator and say in words what I precisely mean. What I precisely mean is you take the sure operator in this super conformal multiplet, which happens to be the super conformal primary. It's the guy at the top. And it's the twisted translated version of that particular operator. And I can compute that j hat z t of 0 is indeed 0 if I pick my factors of 2 and other case and so on correctly. But more interestingly, you will find that j hat z j hat 0 is what? It is 1 over z to the fourth. Now times some funky combination of terms, which I believe is 2k d, where d is the dimension of d, times k 2d plus the do a coccidum number, minus k 2d. This is also k 2d. So this tells me that the norm of this b hat 2 guy is fixed in terms of these sort of canonical numbers in the Chiara algebra. So if you give me the level k and the type of algebra, so I know it's do a coccidum number and dimension, and you also give me the two dimensional, the flavor central charge, or sorry, if you also give me a C central charge, then I just know the norm of this guy. But this is just a norm of a b hat 2 multiplet. It's just a two point function of a b hat 2 multiplet twisted. And in the Chiara algebra, the norm may be positive or negative. But the b hat 2 multiplet upstairs must have a positive norm. That's a unitary theory. So this should also be something like epsilon, but it's a b hat 2. So let me call it m. It's in the 2. So it has 4 r symmetry indices of z z bar u i to u l. And then I have a u a to u d m a b c d of 0, which is a bunch of u's again. Bunch of epsilon i a epsilon j b epsilon l d over x 2 d 8. But some norm. And in conventions that I'm not entirely sure about, but I think it's these conventions, n has to be positive. So unitarity tells me that this is a positive norm. Maybe it's negative. I'm not entirely sure about the science now. But I think with these choice of epsilon's, you basically have to go to a basis where everything is orthogonal. So you have some orthonormal basis. And then you check that the coefficient is indeed positive. But this norm I've computed in terms of the elementary quantities of the theory. And what this tells you is that there is some non-trivial inequality on these coefficients that has to be obeyed in any sort of unitary theory. So sir, you showed that this combination is positive, but is it expected to or? It's not. It's not. Oh, you just find it. So you kind of predicted it to be half-multiple. Does it guarantee that this is positive or? No. So what this leads to is the following constraint. It tells me that if I translate this back to 40 language, remember that c2d is minus 12 times c4d. And k2d is minus c4d over 2. Then I get, if I did this correctly, that there's this inequality over k4d minus 12. And this has to be true in a unitary theory. And it's a non-trivial consequence of the Chiara algebra. In fact, we checked that all the theories that we know, for a large class of theories, we checked, of course, that this thing is inequalities obeyed. We know what happens if it's saturated. If it's saturated, we know that the stress tensor there becomes equal to j squared. So there's a null state in this set of two operators. So j squared minus the stress tensor becomes null when this thing is saturated, which precisely says that the stress tensor, that j squared and t are the same thing, basically. So in other words, the stress tensor of the Chiara algebra is the Sugavara stress tensor. So yes. Yes, OK. The question is if I could, sorry, if you don't know the Chiara algebra, if you don't know the Chiara algebra, the question was if you can find this unitary bound just directly in four dimensions. I am not sure if what it would amount to is, you give me a 40 theory. It has some b hat's multiplets, b hat 1's and b hat 2's. And I would need to know the two point function of the b hat 2 in terms of if, like all I give you is the two point function of the b hat 1's or the four point, not even the four point function, right? I give you the two point function of the b hat 1's and I give you the central charge. And now you have to tell me what the two point function is of the b hat 2 because that is the thing I determined in terms of k and c from the Chiara algebra. No, I used the Chiara algebra to compute the norm of a b hat 2 multiplet, which I could not have done otherwise. Because I computed this j squared two point function from the four point function of single j's by taking the particular limit, right? So I had this four point function which determines the norm of a j squared. It determined this thing. And I don't think I would have been able to determine it otherwise. That's a non-trivial two point function. This thing required crossing symmetry and all of these things to be obeyed and required the mere morphicity. It's not just a short distance thing. So I don't think I would be able to get this number without knowing this. And that's why I needed the Chiara algebra. So we know when it's saturated, that's what I wanted to say. If T sub gavara equals T, so basically this b hat 2 gets zero norm. It decouples. So there is no, in the physical theory, there is no b hat 2 in the singlet in this sector. So only, of course, this is just a subsector. These were just the subsector generated by these guys. And in this sector, there is no b hat 2 in the flavor singlet combination. This b hat 2 disappears. And the suga-vara stress tensor becomes the actual stress tensor. So this also translates, by the way, into some Higgs-Brenge-Karow-Ring relation, which we, yeah, is then a consequence also of the saturation of this equality. So in this case, you get inequality that involves flavor central charge and central charges. More generally, other people have derived bonds like this one, C40 greater than 11 over 30, and other bounds on flavor central charges that don't involve the C central charge, but that depend on the particular algebra that you consider. So it's expressible in terms of some basic quantum, basic numbers in variance of the algebra that they didn't take notes, so I can't reproduce them right now. But they're also K40 inequalities of this type. So these inequalities, I believe, would not be possible to obtain them without knowing this Kyra algebra. But now that you know about Kyra algebras, you can derive them. And these, in some sense, constrain the space of all four-dimensional n equals 2 super conformal fields, both Lagrangian and non-Lagrangian. No matter what theory you have, if it's unitary, if it has a stress tensor, and in this case, if it has flavor, then these bonds that we have derived must be obeyed. So in this sense, this constraints the landscape a little bit further. I think our dream here would be to get even many more constraints, such that maybe in some sense, you can completely classify the space of four-dimensional n equals 2 theories, at least say with small number of generators for the Kyra algebra, just because they are the only consistent Kyra algebras that are compatible with four-dimensional unitary. This is, of course, a very far-fetched dream, but this is sort of a first step in the right direction, where you get at least some non-trivial constraints on the space of four-dimensional theories that hold in full generality. So that's why I think this is a nice result because of its universality. It doesn't need Lagrangian. It doesn't need anything, except 40 n equals 2 theories and basic assumptions there. So that's all I wanted to say about this. There are tons and tons of topics that I could have talked about and didn't. I didn't talk much about the index. I didn't talk much about sort of further ideas about unitarity bounds. I didn't talk about the Higgs branch much and the connection with the Higgs branch as a variety, how you can get this from the Kyra algebra. There's a beautiful story there. I didn't talk about defects. I didn't talk about connections with localization. I didn't talk about connections with holography. I didn't talk about partition functions with defects. And I think that's about the list of topics that people have already written papers on. I didn't talk about six-dimensions. I didn't talk about two-dimensions and et cetera, et cetera. So there are really tons and tons of things that I could have followed up on. But I think since I'm out of time, I'll just stop here. Thank you.